# Homework Help: Electric field of infinitely long parallel wires

1. Oct 26, 2008

### E&M

1. The problem statement, all variables and given/known data
Two infinitely long parallel wires separated by a distance 2d, one carries uniform linear charge density of $$\lambda$$ and the other one carries an uniform linear charge density of -$$\lambda$$, find the electric field at a point distance z away from the middle point of the two wires.

2. Relevant equations

E(left wire) = $$\lambda$$.K .integration (1/r^2). dr
E(right wire) = - $$\lambda$$.K .integration (1/r^2). dr

3. The attempt at a solution
r = (x^2 + d^2 + z^2)
dr = dx
i don't know what should be the limit of integration and i don't know if whatever i did is right or not.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 26, 2008

### tan90

hey, I have attached a figure. I thought this might make your life a little bit easier...i think you have to calculate potential and then calculate E, otherwise it's gonna be hard. I think the x and z components cancel each other. so, you will only have y component of electric field. good luck.

#### Attached Files:

• ###### fig.jpg
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3. Oct 26, 2008

### gabbagabbahey

Do you know the solution for a single infinite wire carrying charge density $\lambda$? If so, you can use the superposition principle and avoid having to carry out any integration.

4. Oct 26, 2008

### E&M

i know the solution for single infinite wire at a point z distance above the center of the wire but since, here we also have to consider y direction, I am not sure how to do it.

5. Oct 26, 2008

### gabbagabbahey

Well, what is the solution "for single infinite wire at a point z distance above the center of the wire"?

6. Oct 26, 2008

### E&M

K. 2 (lambda) / z for the wire with charge density = positive. lambda

7. Oct 26, 2008

### gabbagabbahey

Good, so for that solution z is the distance from the wire....What is the distance from the point (0,0,z) to either of the two wires in the new problem?

8. Oct 26, 2008

### E&M

it is (x^2 + y^2 + z^2) ^.5

9. Oct 26, 2008

### gabbagabbahey

The distance would be the hypotenuse in the following diagram...
k
|
|
|\
| \
|z \
|___\_______>j
d

wouldn't it?

10. Oct 26, 2008

### E&M

if u were asking distance from the center of the wire then it's gonna be (y^2 + z^2) ^.5 where y = d right?

11. Oct 26, 2008

### E&M

so, is the E simply going to be K. 2 (lambda) / (d^2 + z^2)^.5 along the direction of (d^2 + z^2) and then when we consider two wires, z components are going to cancel and we calculate and add the y components?

12. Oct 26, 2008

### E&M

am I anywhere close?

13. Oct 26, 2008

### gabbagabbahey

Yes, the magnitude of E for the wire with +lambda is just K. 2 (lambda) / (d^2 + z^2)^.5. And the magnitude of the other wire is (-)K. 2 (lambda) / (d^2 + z^2)^.5. What are the directions of each of those fields?

14. Oct 26, 2008

### E&M

for the individual wires, their x components cancel and they only have y and z components, but if we consider both of the wires, their z components cancel as well and then there will be only y component left pointing towards the wire with -lambda charge density.

15. Oct 26, 2008

### E&M

am I right?

16. Oct 26, 2008

### gabbagabbahey

Yes, so what is the y-component of E for each wire? (Remember that E points along the hypotenuse in the above diagram)

17. Oct 26, 2008

### E&M

individual y component is going to be K* 2 * (lambda) * cos( angle between hypotenuse and d) / (d^2 + z^2)^.5.
= K* 2 * (lambda) * d / (d^2 + z^2)
Net y component = 2* K* 2 * (lambda) * d / (d^2 + z^2)
????????????

18. Oct 26, 2008

### gabbagabbahey

Looks good to me

$\vec{E}(z)=\frac{4k \lambda d}{z^2+d^2} \hat{y}$

19. Oct 26, 2008

### E&M

is lambda missing in your equation ?

20. Oct 26, 2008

### E&M

thanks a lot

21. Oct 26, 2008

### gabbagabbahey

yes, lambda was missing, but it's fixed now.